How to evaluate the integral $$\int_{0}^{\infty}t^\alpha\mathrm{e}^{-t(1+\mathrm{j}\omega)}\mathrm{d}t$$ for $\alpha>-1$, $\mathrm{j}^2=-1$, and $\omega>0$?
I think it should be equal to $\frac{1}{(1+\mathrm{j}\omega)^{\alpha+1}}\Gamma(\alpha+1)$ with $\Gamma$ the Gamma function defined as $$\Gamma(\alpha+1)=\int_0^{\infty}t^\alpha\mathrm{e}^{-t}\mathrm{d}t.$$ How to show this? Does someone know a reference?
\begin{align} \int_0^\infty t^\alpha e^{-t(1+j\omega)}\mathrm dt &= \sum_{k=0}^\infty \frac{(-j \omega)^k}{k!}\underbrace{\int_{0}^\infty t^{\alpha + k}e^{-t}\mathrm dt}_{=\Gamma(\alpha + k + 1)}\\ &= \sum_{k=0}^\infty \frac{(-j \omega)^k}{k!}(\alpha+k)(\alpha + k - 1)\cdots(\alpha+1)\Gamma(\alpha + 1)\\ &= \Gamma \left(\alpha + 1\right) \underbrace{\sum_{k=0}^\infty\frac{(\alpha + 1) \cdots (\alpha + k)}{k!} (-j\omega)^k}_{=\frac{1}{(1-(-j\omega))^{\alpha+1}}}\\ &= \frac{\Gamma (\alpha + 1)}{(1+j\omega)^{\alpha + 1}} \end{align}